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asambeis [7]
3 years ago
10

To determine the trajectory of the movement of a material point in the xOz plane that is subjected simultaneously to oscillation

s x (t) = A * sin π(t + 1) and y (t) = A * sin (πt + π / 2).
Physics
1 answer:
nasty-shy [4]3 years ago
4 0

We are to show that the given parametric curve is a circle.

The trajectory of a circle with a radius r will satisfy the following relationship:

(x-x_c)^2 + (y-y_c)^2 = r^2

(with (x_c,y_c) being the center point)

We are given the x and y in a parametric form which can be further rewritten (using properties of sin/cos):

x(t) = A\sin \pi(t+1) = A\sin (\pi t + \pi) = -A\sin \pi t\\y(t) = A\sin (\pi t + \frac{\pi}{2}) = A\cos \pi t

Squaring and adding both gives:

x^2(t) + y^2(t) = A^2(-1)^2\sin^2 \pi t + A^2 \cos^2 \pi t = A^2\\\implies x^2 + y^2 = A^2

The last expression shows that the given parametric curve is a circle with the center (0,0) and radius A.


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A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a
Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

x=\sqrt {\frac {2mgh}{k}}

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m

5 0
3 years ago
The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s
alex41 [277]
Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
8 0
4 years ago
which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload
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5 0
3 years ago
An elevator is moving down with an acceleration of 3.36 m/s2.
sergeinik [125]

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

  • From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
  • When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

\sf\longrightarrow Weight = mg - ma \\

\sf\longrightarrow Weight = m ( g - a ) \\

  • Mass of the man = 64.2 kg

\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\

\sf\longrightarrow Weight = 64.2 * 6.44 N\\

\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\

5 0
2 years ago
PLEASE HELP, LOOK AT PHOTO PLZ!!!
PIT_PIT [208]

Answer:

a) 17 km

b) 9 km

Explanation:

The distance is the length of the path.

A to C: 5 km

B to C: 4 km

C to B: 4 km

B to C: 4 km

Total distance = 5 km + 4 km + 4 km + 4 km = 17 km

Displacement is the difference between the starting point and ending point.

Displacement = 9 km − 0 km = 9 km

3 0
3 years ago
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